# How do you solve ln x - ln(x-1) = 2?

Apr 4, 2018

$x = {e}^{2} / \left({e}^{2} - 1\right)$

#### Explanation:

First note that

$\ln x - \ln \left(x - 1\right) = \ln \left(\frac{x}{x - 1}\right)$

So for your problem we can write

$\ln \left(\frac{x}{x - 1}\right) = 2$.

Now take the exponential of both sides

$\frac{x}{x - 1} = {e}^{2}$

(Another way to think of this is if $\ln a = b$, then $a = {e}^{b}$.)

$x = x {e}^{2} - {e}^{2}$

$x = {e}^{2} / \left({e}^{2} - 1\right) = \frac{1}{1 - \frac{1}{e} ^ 2}$